To find the last two digits of $39^{160}$, we need to compute:
$39^{160} \mod 100$

Step 1: Applying Euler’s Theorem

Since $\gcd(39, 100) = 1$, Euler's theorem applies.
Euler’s totient function $\phi(100) = 40$, so:
$39^{40} \equiv 1 \pmod{100}$

Step 2: Simplify the Exponent

$39^{160} = (39^{40})^4$ Using the previous congruence: $(39^{40})^4 \equiv 1^4 \equiv 1 \pmod{100}$

Final Answer:

The last two digits of $39^{160}$ are 01.

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Would you like more details on Euler's theorem or the totient function?

Related Questions:

1. How does Euler’s theorem work, and when can you apply it?
2. What is the totient function $\phi(n)$, and how do you calculate it?
3. How would this problem change if the modulus were different, like 50 or 125?
4. Can you solve $39^{160} \mod 101$ (a prime modulus)?
5. What happens if $\gcd(a, m) \neq 1$ in such problems?

Tip:

For last two digits problems, always think in terms of mod 100 and consider Euler's theorem if the base and 100 are coprime!